A group of research proposals was evaluated by a panel of experts to decide whether or not they were worthy of funding. When these same proposals were submitted to a second independent panel of experts, the decision to fund was reversed in 40% of the cases. If the probability that a proposal is judged worthy of funding by the first panel is 0.3, what are the probabilities of these events

Let A represent the event that the first group approves the proposal.

Let B represent the event that the second group approves the proposal.

We know the following:

P(A) = 0.3

P(B given A) = 1 - 0.3 = 0.7

P(not B given not A) = 1 - 0.3 = 0.7 .

A) P(approved by both groups)

= P(A and B)

= P(A)P(B given A)

= 0.3(0.7)

= 0.21 .

B) P(disapproved by both groups)

= P(not A and not B)

= P(not A)P(not B given not A)

= (1 - 0.3)(0.7)

= 0.49 .

C) This problem might be ambiguous...do you mean exactly one group or at least one group? So, I will use both interpretations.

P(approved by at least one group)

= 1 - P(disapproved by both groups)

= 1 - 0.49

= 0.51 .

P(approved by exactly one group)

= P(second group reverses the first group's decision)

= 0.4 .

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!